Answer:
Explanation:
Assume we have 100g of this substance. That means we would have 20.24g of Cl and 79.76g of Al. Now we can find how many moles of each we have:
[tex]\frac{79.76 \:g}{35.45 \: g/mol}[/tex] = 2.25 mol of chlorine
[tex]\frac{20.24 \: g}{26.98 \: g/mol}[/tex] = 0.750 mol of Al.
To form a integer ratio, do 2.25/0.75 = 2.99999 ~= 3.
So the ratio is essentially Al : Cl => 1 : 3. To the compound is possibly [tex]AlCl_3[/tex].
However, it says it has a molar mass of 266.64 g/mol, and since AlCl3 has a molar mass of 133.32, it must be [tex]Al_2Cl_6[/tex].
Actually this molecule isn't exactly AlCl3 (which is ionic). Al2Cl6 forms a banana bond where Cl acts as a hapto-2 ligand. But that's a bit advanced. All you need to know is X = Al2Cl6
The molecular formula of the compound is Al₂Cl₆
To solve the question given above, we'll begin by obtaining the empirical formula of the compound. This can be obtained as follow:
Aluminum (Al) = 20.24%
Chlorine (Cl) = 79.76%
Al = 20.24%
Cl = 79.76%
Divide by their molar mass
Al = 20.24 / 27 = 0.75
Cl = 79.76 / 35.5 = 2.25
Divide by the smallest
Al = 0.75 / 0.75 = 1
Cl = 2.25 / 0.75 = 3
Thus, the empirical formula of the compound is AlCl₃
Finally, we shall determine the the molecular formula of the compound.
Molar mass of compound = 266.64 g/mol
Empirical formula = AlCl₃
[AlCl₃]ₙ = 266.64
[27 + (3×35.5)]n = 266.64
[27 + 106.5]n = 266.64
133.5n = 266.64
Divide both side by 133.5
n = 266.64 / 133.5
Molecular formula = [AlCl₃]ₙ
Molecular formula = [AlCl₃]₂
Therefore, the molecular formula of the compound is Al₂Cl₆
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